• Jul 17th 2009, 07:04 AM
PaulRS
Let $\displaystyle p$ be a prime such that $\displaystyle p>(n^2+n+k)^2+k$. ( $\displaystyle n,k \in \mathbb{Z}^ +$ )

Consider the sequence: $\displaystyle n^2 ,n^2 + 1,...,\left( {n^2 + n + k} \right)^2 + k$.

Prove that there's a pair $\displaystyle (m,m+k)$ of integers from our sequence, such that $\displaystyle \left( {\tfrac{m} {p}} \right)_L = \left( {\tfrac{{m + k}} {p}} \right)_L = 1$

This is a lovely problem, have fun!. (Happy)

PS: By the way, I'd posted 2 problems like in January, NCA gave a solution for 1, but the other is waiting! (haha, and it is a nice problem too)
• Jul 17th 2009, 08:26 PM
NonCommAlg
Quote:

Originally Posted by PaulRS
Let $\displaystyle p$ be a prime such that $\displaystyle p>(n^2+n+k)^2+k$. ( $\displaystyle n,k \in \mathbb{Z}^ +$ )

Consider the sequence: $\displaystyle n^2 ,n^2 + 1,...,\left( {n^2 + n + k} \right)^2 + k$.

Prove that there's a pair $\displaystyle (m,m+k)$ of integers from our sequence, such that $\displaystyle \left( {\tfrac{m} {p}} \right)_L = \left( {\tfrac{{m + k}} {p}} \right)_L = 1$

This is a lovely problem, have fun!. (Happy)

PS: By the way, I'd posted 2 problems like in January, NCA gave a solution for 1, but the other is waiting! (haha, and it is a nice problem too)

if $\displaystyle \left(\frac{n^2+k}{p} \right)=1,$ choose $\displaystyle m=n^2$ and if $\displaystyle \left(\frac{(n+1)^2+k}{p} \right)=1,$ then choose $\displaystyle m=(n+1)^2.$ so we may assume that $\displaystyle \left(\frac{n^2+k}{p} \right)=\left(\frac{(n+1)^2+k}{p} \right)=-1.$

now we have this cute identity: $\displaystyle (n^2+k)((n+1)^2+k)=(n^2+n+k)^2+k.$ therefore: $\displaystyle \left(\frac{(n^2+n+k)^2+k}{p} \right)=1$ and we choose $\displaystyle m=(n^2+n+k)^2. \ \Box$

Remark: to put the above result in simple words: for any positive integers $\displaystyle n,k$ and any prime number $\displaystyle p > (n^2+n+k)^2+k,$ at least one of the numbers

$\displaystyle n^2+k, \ (n+1)^2+k, \ (n^2+n+k)^2+k$ is a quadratic residue modulo $\displaystyle p.$